Question

a bag of m&ms has 3 red, 4 green, 5 blue, and 6 yellow m&ms. suppose you randomly select two m&ms from the bag one at a time with replacing the first m&m. let a = first m&m is green and b = second m&m is green. find the following probabilities. (write your answers as fractions.) a) p(a) = b) p(b | a) = c) p(a and b) = question help: written example message instructor

Explanation:

Step1: Calculate total number of M&M's

The total number of M&M's is $3 + 4+5 + 6=18$.

Step2: Calculate P(A)

The probability that the first M&M is green is the number of green M&M's divided by the total number of M&M's. So $P(A)=\frac{4}{18}=\frac{2}{9}$.

Step3: Determine P(B|A)

Since we are sampling with replacement, the probability that the second M&M is green given that the first one is green is the same as the probability of picking a green M&M on any single - draw. So $P(B|A)=\frac{4}{18}=\frac{2}{9}$.

Step4: Calculate P(A and B)

For independent events (since we are sampling with replacement, A and B are independent), $P(A\cap B)=P(A)\times P(B|A)$. Substituting the values we found, $P(A\cap B)=\frac{2}{9}\times\frac{2}{9}=\frac{4}{81}$.

Answer:

a) $\frac{2}{9}$
b) $\frac{2}{9}$
c) $\frac{4}{81}$